Illustrate how sodium sulfide (Na₂S) is formed.
Sodium sulfide (Na₂S) is formed by the transfer of electrons from sodium atoms to a sulfur atom.
A sodium atom loses its one valence electron to become a sodium cation (Na⁺). Since two sodium atoms are needed, each loses one electron, forming two Na⁺ ions.
A sulfur atom has six electrons in its outer shell and needs to gain two electrons to complete its octet. When it gains two electrons (one from each sodium atom), it acquires two units of negative charge and becomes a sulfide anion (S²⁻).
The two Na⁺ ions and one S²⁻ ion are held together by electrostatic forces of attraction to form sodium sulfide (Na₂S).
This can be represented as: 2Na → 2Na⁺ + 2e⁻ S + 2e⁻ → S²⁻ 2Na⁺ + S²⁻ → Na₂S
Explanation
The question asks students to illustrate the formation of Na₂S through electron transfer, similar to how NaCl formation is shown in Fig. 9.13. From the context: sodium loses one valence electron to become Na⁺ (11 protons, 10 electrons), and sulfur has six valence electrons and gains two to become S²⁻. Since each Na gives only 1 electron but S needs 2, two Na atoms are required. The pattern follows the same logic as Pause and Ponder Q13, where one Mg²⁺ combines with two Cl⁻ ions — here, two Na⁺ ions combine with one S²⁻ ion.
Solution Steps
Step 1: Identify that sodium (Na) loses 1 electron to form Na⁺ cation
Step 2: Identify that sulfur (S) gains 2 electrons to form S²⁻ anion
Step 3: Determine that 2 Na atoms are needed to provide 2 electrons for 1 S atom
Step 4: Show the electron transfer: 2Na → 2Na⁺ + 2e⁻ and S + 2e⁻ → S²⁻
Step 5: Combine the ions: 2Na⁺ + S²⁻ → Na₂S